Solution:

119

Explanation:
Let the minimum number of chocolates = C
When divided by 2, remainder = 1
When divided by 3, remainder = 2
When divided by 4, remainder = 3
When divided by 5, remainder = 4
When divided by 6, remainder = 5
When divided by 7, remainder = 0

Therefore, C is divisible by 7.

Whenever C is divided by any number less than 7, the remainder is 1 less than the divisor.
This implies that, C + 1 is the LCM of 2, 3, 4, 5 and 6.

Now LCM of 2, 3, 4, 5 and 6 = 60
But 60 - 1 = 59 is not divisible by 7

60 * 2 = 120
120 - 1 = 119 which is divisible by 7

Therefore, C + 1 = 120 or C = 119

Number of minimum number of chocolates in the box = 119

Solution:

The Bangladeshi seaman is the thief. This is because the Japanese flag cannot be identified as upside down as it is the same both ways. Thus he is just trying to escape the situation.

Solution:

2 cakes

At each toll bridge you are required to give half of your cakes, and you receive one back. Which leaves you with 2 cakes after every bridge.

Solution:

Are you Thorgan

Explanation:
Below will be the reply of three brothers:
Eden: No (speaks truth)
Thorgan: No (speaks lie)
Kylian: Yes (speaks lie)

Therefore if its Kylian then only you will get a reply as "Yes" and then you ask for your money.

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93

Solution:

The ball will take infinite bounces before it is stopped. As per the question it will keep on bouncing half way up every time it hits the ground. But gravity will play its part at certain point of time which will make it stop.

Solution:

2, 3, 10 and 10
To maintain the difference of eight and keeping in mind the third and fourth hint, the first and third cards can be 2 and 10 or 10 and 2. But keeping in mind the first hint, we know that the first card is 2 and the third is 10. Similarly, if we see the fifth hint, the difference between the second and fourth card should be 7. Thus we know that the second card is 3 and the fourth card is 10.

Solution:

111111111

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

Solution:

They were playing tennis.

Tiger Phil
30 - 30
40 - 30 (Phil hit a bad shot)
Game won - 30 (Tiger hit an awesome shot)

Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

Never

Explanation:
In the question, the distance of x miles is given at a particular moment. The middle car is running at that time at 50 MPH. The distance will keep increasing with time as three are running at different speeds. It will increase by x miles every hour till a certain point of time and then it will start decreasing till the fastest car meets the slowest again.

As per question, all the cars stop then. Thus, the distance will never be x miles again.